# How do you solve 3x^2-16x+5<=0 by algebraically?

May 12, 2018

The solution is $x \in \left[\frac{1}{3} , 5\right]$

#### Explanation:

First factorise the equation

$3 {x}^{2} - 16 x + 5 = \left(3 x - 1\right) \left(x - 5\right) \le 0$

$\left(3 x - 1\right) \left(x - 5\right) = 0$

when

$\left\{\begin{matrix}3 x - 1 = 0 \\ x - 5 = 0\end{matrix}\right.$

$\left\{\begin{matrix}x = \frac{1}{3} \\ x = 5\end{matrix}\right.$

Let $f \left(x\right) = \left(3 x - 1\right) \left(x - 5\right)$

Build a sign chart

$\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a a a a}$$\frac{1}{3}$$\textcolor{w h i t e}{a a a a a a a a}$$5$$\textcolor{w h i t e}{a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$3 x - 1$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$0$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$x - 5$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$color(white)(aaaaa)-$\textcolor{w h i t e}{a a}$$0$$\textcolor{w h i t e}{a a}$$+$

$\textcolor{w h i t e}{a a a a}$$f \left(x\right)$$\textcolor{w h i t e}{a a a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$0$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a}$$0$$\textcolor{w h i t e}{a a}$$+$

Therefore,

$f \left(x\right) \le 0$ when $x \in \left[\frac{1}{3} , 5\right]$